Question

2 log 3x - log 9 = 1
Can anyone explain how to solve this please?

Answer 1

The first step is remembering a few log rules: \[a*\log b = \log b ^{a}\] and \[\log a - \log b = \log(a/b)\]
Once you remember these two rules, you can simplify the equation:\[\log ((3x)^{2}/9) = 1\]
Next, simplify the part inside of the log and you will end up with: \[\log x ^{2} = 1\]
To solve for x, use another log rule: \[10^{\log a} = a\]
Thus (because you have to do the same thing to both sides), \[10^{\log x ^{2}} = 10^{1}\]
And finally:\[x ^{2} = 10, x = \sqrt{10} or -\sqrt{10}\]
BUT since in the original equation you have log (3x), x CANNOT be negative (another log rule). So the answer is:\[x = \sqrt{10}\]

Answer 2

Wow thanks for the help! I was really confused about that! 😄